Current through an AC series L-C-R circuit is 2 A if operated at resonance frequency , and 1 A if operated at 50% less than resonant frequency . The current (in A) if the frequency is 100% more than the resonant frequency , is
A.`sqrt 2` B.1 C.`sqrt 3 ` D.Data insufficient

Options

A. `sqrt 2`
B. 1
C. `sqrt 3 `
D. Data insufficient

Explanation

To solve the problem, we need to analyze the behavior of the current in an AC series L-C-R circuit at different frequencies, specifically at resonance frequency and at frequencies that are 50% less and 100% more than the resonance frequency.

### Step-by-Step Solution:

1. **Identify the Resonance Condition**:
At resonance frequency (ω₀), the inductive reactance (X_L) equals the capacitive reactance (X_C). This means:
\[ X_L = X_C \]

2. **Current at Resonance Frequency**:
The current through the circuit at resonance frequency is given as:
\[ I_0 = 2 \, \text{A} \]

3. **Current at 50% Less than Resonance Frequency**:
When the frequency is 50% less than the resonance frequency, the new frequency (ω') is:
\[ \omega' = \frac{\omega_0}{2} \]
At this frequency:
- The inductive reactance becomes:
\[ X_L' = \frac{X_L}{2} \]
- The capacitive reactance becomes:
\[ X_C' = 2X_C \]

4. **Calculate the Total Impedance at 50% Less Frequency**:
The total impedance (Z') at this frequency is given by:
\[ Z' = R + j(X_C' - X_L') = R + j(2X_C - \frac{X_L}{2}) \]
Since we know that at resonance \(X_C = X_L\), we can substitute \(X_C\) with \(X_L\):
\[ Z' = R + j(2X_L - \frac{X_L}{2}) = R + j(\frac{4X_L}{2} - \frac{X_L}{2}) = R + j(\frac{3X_L}{2}) \]

5. **Current at 50% Less Frequency**:
The current at this frequency is given as:
\[ I' = 1 \, \text{A} \]
Using Ohm's law:
\[ I' = \frac{V}{Z'} \]

6. **Current at 100% More than Resonance Frequency**:
When the frequency is 100% more than the resonance frequency, the new frequency (ω'') is:
\[ \omega'' = 2\omega_0 \]
At this frequency:
- The inductive reactance becomes:
\[ X_L'' = 2X_L \]
- The capacitive reactance becomes:
\[ X_C'' = \frac{X_C}{2} \]

7. **Calculate the Total Impedance at 100% More Frequency**:
The total impedance (Z'') at this frequency is:
\[ Z'' = R + j(X_C'' - X_L'') = R + j(\frac{X_C}{2} - 2X_L) \]
Again substituting \(X_C\) with \(X_L\):
\[ Z'' = R + j(\frac{X_L}{2} - 2X_L) = R + j(\frac{X_L - 4X_L}{2}) = R + j(-\frac{3X_L}{2}) \]

8. **Determine the Current at 100% More Frequency**:
The current at this frequency can be expressed as:
\[ I'' = \frac{V}{Z''} \]
Since the impedance changes but the voltage remains constant, we can analyze the ratio of the currents:
\[ \frac{I'}{I_0} = \frac{Z}{Z'} \quad \text{and} \quad \frac{I''}{I_0} = \frac{Z}{Z''} \]
Given that the impedance at 50% less frequency resulted in a current of 1 A, we can deduce that the current at 100% more frequency will also yield a current of 1 A due to the symmetry in the behavior of the circuit.

### Final Answer:
The current when the frequency is 100% more than the resonant frequency is:
\[ \text{Current} = 1 \, \text{A} \]